Residue theorem application

Question

How could we use the Residue theorem to calculate the following integral:

$$\int_0^{2\pi} \frac{1}{1-2p\cos{x} + p^2} dx$$

where $p$ is a real constant, such that $p\in ]0,1[$

Thank you!

Answer 1

$$1-2p\cos x+p^2= (1-pe^{i\theta})(1-pe^{-i\theta})$$ hence by setting $z=e^{i\theta}$ it follows that: $$\int_{0}^{2\pi}\frac{dx}{1-2p\cos x+p^2} = -i\oint\frac{dz}{(1-pz)(z-p)} $$ where the last countour integral is over the unit circle. Since $p\in(0,1)$, $$ -i\oint\frac{dz}{(1-pz)(z-p)} = 2\pi \operatorname{Res}\left(\frac{1}{(1-pz)(z-p)},z=p\right)=\color{red}{\frac{2\pi}{1-p^2}}$$ by the residue theorem.

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With a real-analytic approach: $$\begin{eqnarray*}\int_{0}^{2\pi}\frac{dx}{1-2p\cos x+p^2}&=&2\int_{0}^{\pi}\frac{dx}{(1+p)^2-4p\cos^2 x}\\&=&4\int_{0}^{+\infty}\frac{dt}{(1+p)^2(1+t^2)-4p}\\&=&\frac{4}{|1-p^2|}\int_{0}^{+\infty}\frac{du}{1+u^2}\\&=&\frac{2\pi}{|1-p^2|}.\end{eqnarray*}$$